Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
+1 vote
32.5k views
in Physics by (130k points)
edited by

In accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution around the sun in an orbit of radius 1.5 × 1011 m with orbital speed 3× 104 m/s. (Mass of earth = 6.0 × 1024 kg.)

1 Answer

+1 vote
by (93.9k points)
selected by
 
Best answer

Radius of the orbit of the Earth around the Sun, r = 1.5 × 1011 m
Orbital speed of the Earth, ν = 3 × 104 m/s
Mass of the Earth, m = 6.0 × 1024 kg
According to Bohr’s model, angular momentum is quantized and given as:

Where,
h = Planck’s constant = 6.62 × 10−34 Js
n = Quantum number

Hence, the quanta number that characterizes the Earth’ revolution is 2.6 × 1074.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...